I How Do You Calculate Equilibrium Conditions After Mixing Two Different Gases?

  • I
  • Thread starter Thread starter jaumzaum
  • Start date Start date
  • Tags Tags
    Gases Mixing
AI Thread Summary
To calculate the equilibrium conditions after mixing two different gases in adiabatic containers, one must use the conservation of energy alongside the ideal gas law. The internal energy of a monoatomic gas, which depends solely on temperature, can be expressed as E = (3/2)nRT. Since the process is irreversible, calculating the work done by the gases can be complex, and one cannot assume constant entropy. Clarification on whether the gases are the same and if thermal equilibration is allowed is also essential for accurate calculations. Understanding these principles is crucial for determining the final pressure and temperature after mixing.
jaumzaum
Messages
433
Reaction score
33
Consider that I have 2 adiabatic containers, one with a monoatomic gas at pressure P1, volume V1 and temperature T1 and another with pressure P2, volume V2, temperature T2. If I open a valvule and mix the two gases, how do I calculate the equilibrium pressure and temperature?

I know the final quantity of moles, and the final volume, so I need a system of 2 eq to calculate the temperature and pressure. One is the Clapeyron Law, what is the other? I mean, what variable remains constant in this system?

If I try to solve it using conservation of energy:
dU1+dW1+dU2+dW2=0

But how would I calculate the work done by the gases, as this is an irreversible transformation?Thanks!
 
Science news on Phys.org
Can I say the entropy is constant?
 
You can definitely not assume that entropy is constant. Furthermore, the Clapeyron Law apparently is apparently just another name for the ideal gas law. I never heard the name Clapeyron Law before...

The other equation you need is just the conservation of energy. The internal energy of a monatomic gas is:
$$E = \frac{3}{2}nRT$$
This only depends on temperature. So with this you can compute the temperature, then with the ideal gas law you can compute the pressure.
 
jaumzaum said:
Consider that I have 2 adiabatic containers, one with a monoatomic gas at pressure P1, volume V1 and temperature T1 and another with pressure P2, volume V2, temperature T2. If I open a valvule and mix the two gases, how do I calculate the equilibrium pressure and temperature?

I know the final quantity of moles, and the final volume, so I need a system of 2 eq to calculate the temperature and pressure. One is the Clapeyron Law, what is the other? I mean, what variable remains constant in this system?

If I try to solve it using conservation of energy:
dU1+dW1+dU2+dW2=0

But how would I calculate the work done by the gases, as this is an irreversible transformation?Thanks!
Is the gas the same in the two containers? Are you allowing the two containers to equilibrate with one another thermally?
 

Thread 'Basic thermal energy transfer physics: Insulation value in open airways vs sealed airways'

Been around 40 years since I took basic physics in college and while I remember doing some examples of insulation values / energy conduction, I doubt I could to the math now even if I could find the formulas. I have some some corrugated plastic sheet (think of the plastic signs you see on the side of the road) that is used in bee hives. Also have some used in a green house though a bit different in dimensions than this example but the general approach should still apply. Typically, both...
View full post »

Thread 'Using battery energy to heat a battery…'

Problem: You’re an Uber driver with a Tesla Model 3. Today’s low: 30F, high: 65F. You want to reach a USD$ profit target in the least number of hours, but your choices could have added cost. Do you preheat the battery only when you are headed to the charging station (to increase the charging rate by warming the battery — however the battery might not be “warm enough” when your reach the charger and thus slower charging rates), or do you always “navigate to the charger” the entire day (which...
View full post »
Thread 'Is Callen right in claiming dQ=TdS for all quasi-static processes?'

Thread 'Is Callen right in claiming dQ=TdS for all quasi-static processes?'

Hello! I am currently reading the second edition of Callen's Thermodynamics and an Introduction to Thermostatistics, and I have a question regarding Callen's definition of quasi-static. On page 96, Callen says: Another way of characterizing Callen's definition is that a process is quasi-static if it traces out a continuous curve in the system's configuration space. So far it's all well and good. A little later, Callen claims that the identification of $$TdS$$ as the heat transfer is only...
View full post »

Similar threads

I Movable Wall in an Adiabatic System: Same Final Temperature?
Replies
8
Views
1K
I Why work is PdV and not (P+dP)dV in an isothermal process?
Replies
6
Views
216
I Focus Problem for Entropy Change in Irreversible Adiabatic Process
2
Replies
60
Views
9K
Mixing two ideal gases with different V, T at constant pressure
Replies
16
Views
4K
Chemistry How does it work to mix two gases reversibly in this device?
Replies
1
Views
2K
Mixing two gases in an isolated system and calculating final p and T
Replies
4
Views
2K
I How to calculate the mass of gas in a tank?
3
Replies
109
Views
7K
I Irreversible adiabatic processes & entropy change (clarification needed)
Replies
22
Views
5K
B Energy seems to disappear when gas is compressed?
Replies
3
Views
2K
I Graphical difference between adiabatic and isothermal processes.
Replies
1
Views
3K

Hot Threads

Recent Insights

Back
Top